Shear reinforcement in reinforced concrete floor systems



Nov. 8, 1966 B. GERSOVITZ 3,28

SHEAR REINFORCEMENT IN REINFORCED CONCRETE FLOOR SYSTEMS Filed May 10,1963 5 Sheets-Sheet 1 Lgh a (c+ 03} M56 00i 6/- Y Benjamin GersovhzmfiW%# ATTORNEYS Nov. 8, 1966 sovrrz 3,283,458

SHEAR REINFORCEMENT IN REINFORCED CONCRETE FLOOR SYSTEMS Filed May 10,1963 5 Sheets-Sheet 2 INVENTOR Benjamin Gersovitz ATTORNEYS Nov. 8, 1966B. GERSOVITZ 3,283,458

SHEAR REINFORCEMENT IN REINFORCED CONCRETE FLOOR SYSTEMS Filed May 10,1963 5 Sheets-Sheet 5 INVENTOR Benjamin Gersovirz EMMMEM ATTORNEYS B.GERSOVlTZ Nov. 8, 1966 SHEAR REINFORCEMENT IN REINFORCED CONCRETE FLOORSYSTEMS Filed May 10, 1963 5 Sheets-Sheet 4 INVENTOR Benjamin Gersovirzwfi wgw ATTORNEYS SHEAR REINFORCEMENT IN REINFORCED CONCRETE FLOORSYSTEMS Filed May 10, 1963 B. GERSOVITZ Nov. 8, 1966 5 Sheets-Sheet 5INVENTOR Benjamin Gersovirz wgw ATTORNEYS columns merge directly withthe ceiling.

United States Patent 1 Claim. (61. 52-260) This is acontinuation-in-part of applicants U.S. application S.N. 790,960, filedFebruary 3, 1959, and now abandoned.

This invention relates to a reticulated structure for providing shearreinforcement in reinforced concrete floors and footings and inparticular is concerned with a prefabricated shearhead for providingshear reinforcement above columns supporting fiat plate floors.

Various types of reinforced concrete floor systems are in current use bythe build-ing industry and are generally well known in the art. Asuitable reference for the various types of floors, their names and usesis the brochure entitled Reinforced Concrete Floor Systems published bythe Portland Cement Association. In this brochure, reference is made toflat slab floor systems and it is in connection with a particularvariety of flat slab floor system known as the flat plate floor that thepresent invention finds particular, though not unique, employment.

Modern design methods for reinforced concrete building, using girderlessfloors base, the safety of such structures on their ultimate strength;and a theory based on a consideration of the ultimate flexural capacityof the slabs used in the floors and footings of such buildings, known asthe yield line theory, has been evolved. It is a basic assumption ofthese design methods, and a premise of this theory, that the slabs usedin a construction method will not fail in shear before the full flexuralcapacity of slab is reached. Therefore it becomes mandatroy for thesafety of such buildings that means are provided to ensure that shearfailure cannot occur. Since the maximum shear stress is generallyreached in the neighborhood of the columns supporting the floors itfollows that such shear reinforcement must primarily be provided atthese columns.

One known method of providing such shear reinforcement is by the use ofdrop panels and column capitals, and a floor system utilizing these twotypes of reinforcement is known as the Flat-Sla system. Drop panels arethickened portions of the slabs around the columns and act to provideincreased cross-sectional area and depth to resist negative moments andshears. Column capitals are flared shoulders at the tops of columns,often being used in conjunction with drop panels, and are effective toreduce the slab span between columns. The use of the drop panels andcolumn capitals though able to provide the required measure of shearreinforcement is undesirable fora multiplicity of reasons. They areunsightly objects and restrict the freedom of the architect in planningthe layout of the interior of the building; they project below thebottom of the slab and hence interfere with any mechanical workdepending from the ceiling. It would therefore be extremely desirable toeliminate the use of drop panels and column capitals without increasingthe danger of shear failure.

The type of flat slab floor'system which dispenses with the use of droppanels and column capitals is, by definition, known as flat plate floorsystem and in it the As well as overcoming the drawbacks of the abovesystem, the fiat plat floor system is aesthetically purer and moreeconomical than a flat slab floor system as it also permits a "icereduction in the height of the stories and the lengths of verticalconstruction including columns, walls, etc.

A search has, therefore, been conducted for some time by the reinforcedconcrete industry for some means of solving the problem of providingshear reinforcement above columns in flat plate floor systems. Onesolution is offered in the R/C Bulletin entitled ShearheadReinfol-cement for Flat Pilate Floors, published by the Portland CementAssociation which consists of disposing concentrically around the columna number of V-shaped stirrups which provide an area of reinforcementsurrounding the column. Though such .a type of shearhead shouldapparently afford suflicient protection against shear failure, thestirrups do not seem to have sufiicient anchorage as samples testedstill failed in shear, with crushing taking place at the upper and loweredges of the stirrups.

Another attempt at solving the problem of providing shear reinforcementin flat plate floor systems is described in the article entitledShearing Strength of Reinforced Concrete Slabs, by Richard Elstner andEdwind Hognestad which appears in the July, 1956 journal of the AmericanConcrete Institute. Here in series IX of the tests conducted by theabove researchers there is disclosed an embryonic form of the presentinvention in which steel bars are laid out in a cruciform pattern abovethe column. It was the apparent endeavour of these researchers toincrease the ultimate shear strength of the floor in the neighbourhoodof the column to the point where the slab would fail fiexurally, i.e. incompression rather than in shear. They were unsuccessful in theirattempt for though they managed to increase the ultimate load capacityof the slab system by 30%, failure still oocured in shear in all samplestested, and the inference to be drawn from this article is that theresearchers did not consider it practical to use steel bars in areticulated pattern to obtain suflicient shear reinforcement so that theslabs would fail in compression rather than in shear.

The shearhead disclosed herein, although retaining some of the moreelemental features of the structure described in the above article,increases the ultimate load capacity by 75% and failure occurs incompression not in shear, that is the flexural capacity of the slabsystem is fully developed and the problem of providing sufficient shearreinforcement for flat plate floor systems is thereby solved which theabove researchers were unable to do.

According to the present invention there is provided a reinforcedconcrete structure comprising a fiat plate floor having an upper andlower surface, at least one supporting column, and shear reinforcementembedded in said concrete adjacent said column, said reinforcementcomprising a shear head having a first set of reinforcing membersconsisting of a first series of bars arranged substantially parallel inequally spaced side-byside relationship and a second series of similarlyarranged bars at right angles thereto, the bars in each series having anupper horizontal central portion substantially parallel to and adjacentthe upper surface of said floor and opposed horizontal anchoringportions interconnected to said central portion by a pair of downwardly,outwardly inclined portion, said inclined portions of each series lyingin a plane angularly disposed to the axis of the respecparent from thefollowing description, when taken in conjunction with the accompanyingdrawings in which:

FIGURE 1 shows diagrammatically the primary shear stress pattern in anunreinforced flat plate floor.

FIGURE 2 shows the effect on this pattern of intro- I ducing an inclinedreinforcement bar which intersects this primary pattern. FIGURE 2a showsa detail of a suitable reinforcement bar.

FIGURE 3 illustrates the etfect of introducing further reinforcementbars to intersect the secondary shear stress pattern.

FIGURE 4 is the structure of FIGURE 3 presented in a graphical form.

FIGURE 5 depicts a partial plan view of the underside of a flat platefloor supported by square columns.

FIGURES 6, 7, and 8 are used to illustrate the extent to which shearreinforcement must be provided around the top of the column.

FIGURE 9 shows a plan view of the prefabricated shearhead of the presentinvention.

FIGURE 10 shows an end elevation of this shearhead and FIGURE 11 is aperspective view of this shearhead as it would be placed above a column.

FIGURE 12 shows an alternative embodiment of this shearhead when usedabove a column which is situated along the outer wall.

Before embarking upon a description of the preferred shear reinforcementstructure of the present invention it is necessary to describe in somedetail the principle underlying its method of operation.

As is well known the shear stress in a flat slab floor system reaches amaximum value at the head of each column and decreases to a minimum at apoint most remote from such columns. The planes of maximum stress at thehead of a column commence at the junctions of the column with theunderside or compression face of the floor and then proceed outwardlyfrom the column through the floor, at an angle which is approximately 45degrees, to the upper or tension face of the floor.

This is shown in FIGURE 1 where the shear pattern 8.

commences at the junction of the column C with the flat plate floor Fand proceeds outwardly at an angle of 45 degrees. The linear distancebetween opposed points on the upper or compression surface Where theshear line S emerges isdenoted by I. Beyond the two boundary pointswhere the shear line emerges the shear stress decreases as the distancefrom the column increases.

Without reinforcement it is quite easily possible for the shear stressinduced at the head of the column to exceed the shear capacity of theconcrete material forming the flat plate floor and therefore the floorwould fail in shear at the line S of primary maximum shear stress. If,however, reinforcement consisting of a first set of reinforcementmembers is provided and arranged in a grid like pattern symmetricallywith respect to the column and if these steel bars each has inclinedportions which intersect this maximum shear line S, then this primarymaximum shear line will be'moved outwardly from the column a certaindistance to form a secondary shear stress pattern S which starts at thelower end of the inclined portion of the bar and then proceeds up-,ward, at an angle of 45 degrees, to the tension face of the concretefloor. As previously noted, the shear stress reduces as the distancefrom the column increases, accordingly, the new maximum shear stress inthe floor, which is now the secondary shear stress along line 5;, may ormay not fall below that value which can be safely home by the concrete.Such a single layer of reinforcement is shown in FIGURE 2 where the flatplate floor F is supported on column C in themanner described forFIGURE 1. In the event the secondary shear stress along S cannot beborne by the concrete, 3, second setof reinforcing members may be added.The second set. of reinforcing members is likewise arranged symmetricaLDue to the action of bar R, in offsetting the shear line,

the distance between opposed points on either side of the column wherethe shear line breaks through the upper surface of the floor, has nowincreased from the value I, shown in FIGURE 1, to a new greater value 1and, of course, the shear stresses in the floor at this greater distanceare reduced from those which occurred without reinforcement.

A suitable reinforcement bar R is shown in detail in FIGURE 2a. This barhas two inclined portions 2 disposed on either side of a central portion1, and it is obvious that by varying the length of this central portionthe distance apart of inclined portions 2 can be varied so that theywill intersect the lines of maximum shear stress. Extending outwardlybeyond the inclined portions 2 are end portions 3, each of which isshown terminating in an anchoring hook 4 which forms a useful though'notessential adjunct to the bar. The inclined portion subtends an obtuseangle -0 with the central portion and varying this angle obviouslycontrols the horizontal extent of the inclined portion. In use in aconcrete floor, the central portion and the two end portions, anchor theinclined portion and enable it to resist the shear stress. Though theseanchoring portions are shown here in their preferred flat parallelforms, it should be remembered that their function is to anchor theopposite ends of the.

inclined portion.

If, as is generally the case, the reinforcement provided by the singlebar of FIGURE 2 is not sufficient to reduce the shear stress to a valuewhich can be borne by the concrete material which constitutes the floor,then, as previously mentioned, it becomes necessary to provide othersteel reinforcing bars across the secondary pattemof maximum shearstress shown inFIGURE 2. This further reinforcement is shown in FIGURE3, where once again, a flat plate floor F is supported upon a column C,

and a first set of reinforcing members is provided. A

second set of reinforcing is introduced wherein the bars 7 have. alonger central portion than in the first set so that mary and secondaryplanes of maximum shear stress are suflicient to provide adequate shearreinforcement but in certain cases it 'will become necessary to providea third or even fourth set of shear reinforcement members wherein theinclined portion of the bars are positioned further from the column soas to. intersect further shear planes.

It will thus be appreciated that the basic object of this i invention isto provide reinforcement for a column supported fiat plate floor whereinsuch reinforcement includes rods having inclined portions symmetricallyarranged about the column, the inclined portions in each series be-.

ing in a plan inclined with respect to the column and forming thefrustrum surface of a regular pyramid, and in the usual type ofconstruction where the column is square sided this frustrum surface willbe that of a regular square pyramid. This reinforcement must of coursebe substantially uniform and continuous through this surface, andaccording to this invention the necessary reinforcement is supplied bythe inclined portions of an appropriate number of reinforcement barswhich extend upwardly in each frustrum surface. The reinforcement planarsurfaces of each set are positioned symmetrically around the column sothat the surface of one set will intersect the plane of maximum primaryshear stress, and that the second set will intersect the plane ofmaximum secondary shear stress. Anchorange for each inclined portion isprovided by forming two anchoring portions on the reinforcement barextending inwardly and outwardly respectively with respect to the columninto the two anchoring regions, and which, when the concrete floormaterial is poured, firmly anchor the inclined portions. In thepreferred embodiment of the present invention for square sided columns,the anchoring portions are made flat and parallel, and the inwardlyextending anchoring portions of opposed inclined portions in opposedfrustrum surfaces are interconnected to form a common flat centralportion; this latter type of reinforcement bar structure being inessence that shown in FIGURE 2a.

Shear reinforcement is provided in this manner, outwardly from thecolumn in all directions where the shear stress exists till a point isreached where the shear stress is reduced to a value which can be borneby the concrete. Alternatively, the problem can be visualized as a needfor increasing the cross sectional area concentric with the column overwhich the shear force is applied, by extending this concentric perimeteroutwardly from the column until a point is reached where the maximumshear stress, which is equal to the total load on the column divided bythe effective cross sectional shear area, is reduced to a value whichcan be comfortably borne by the concrete material used in the floor orfooting.

It becomes necessary, therefore, in considering the type ofreinforcement structure and the area over which such reinforcement mustbe applied to determine at what point, outwardly of the column, theshear stress has been reduced to an acceptable value.

The first consideration is that it must be known what value of shearstress can suitably be borne by the concrete material which constitutesthe floor. This has been set by the American Concrete Institute atSection 807 of the AC1 code at three percent of the ultimate strength ofthe concrete 0.03 fc, where fc is the twenty-eight day ultimate strengthof the concrete. In a typical case of concrete having a 28 day strength,at 3000 lbs., per square inch, the value of the shear stress which couldcomfortably be borne by the concrete is therefore given as 0.03 x3000that is, 90 lbs. per square inch.

The next step is to determine, from a calculation based on the otherknown factors, at what point the shear stress, outwardly of the column,is reduced to this value of 0.03 is. The value of the shear stress isgiven by:

where P is the total shear load borne by the column b is taken equal tothe perimeter of an area concentric with the column at that distancefrom the column at which it is desired to determine the value of shearstress. The factor d represents the distance from the compressionsurface of the floor to the centroid of the reinforcing steel, and j isa standard factor and is equal to the ratio of the distance between thecentroid compression and the centroid of tension to the depth d, and isgenerally taken as being approximately .875.

The manner in which these various factors are calculated so that b, theunknown dimension, can be found from the above equation is as follows:

In FIGURE 5 is shown a typical section of a floor layout in which thecolumns are uniformly spaced at a distance n apart. It will be seen thatthe area supported by each column is equal to I1 From this area thecolumn receives a live load due to the weight of materials. etc. placedupon the floor and a reasonable value can be assumed for this quantity,and in addition it receives a dead load due to the weight of the floorconstruction, which will vary with the thickness of the floor and thetypes of material used in its construction. Both the live and dead loadsare conveniently expressed as so many pounds per square foot and thetotal load P on any column is given by the sum of the live and dead loadunits multiplied by the factor n where n is in feet.

As mentioned above, at is the distance from the compression surface tothe centroid of the reinforcing steel floor and this is shown in FIGURE6 where this steel T is located at a depth d below the upper surface ofthe floor and overall thickness t. Since, as is common practice, thesteel is located at a distance of approximately 1.5 inches, above thelower surface of the floor, then d is given by t1.5 inches and for thecase of a six inch floor would therefore be 6-1.5 i.e. 4.5 inches.

The significance of b is shown in FIGURE 7 which shows a plan view of asection of floor supported upon a square column having a side dimensionc. The peripheral area b has been defined as the perimeter of an areaconcentric with the column and this area will thus take the shape of asquare concentric with the column whose sides as shown are of length p,and b therefore given by 4p, where p is the distance between similarpoints on either side of the column -at which the shear stress is to becomputed. Thus to calculate the dimension of the square over which shearreinforcement must be provided in order to reduce the shear stress to anacceptable level it is necessary to first determine b and from this thevalue of p the length of the side of the square.

If, in the above equation for the value of the shear stress the generalvalue V of the shear stress is replaced by the specific value of theshear stress which can be borne by the floor which as stated, isgenerally taken as 0.03 fc then we may write This can be solved for bsince all the other factors are known for any specific fi-oorconfiguration and so the perimeter of the area of reinforcement isdetermined.

As an example of the manner in which the above formula can be applied,consider the case of a flat plate floor 3000 lbs/sq. inch 28 daystrength concrete composed of 6 inch slabs which is borne upon 14 inchsquare columns set apart at 20 foot intervals. If the live load on thefloor is taken at lbs. sq. ft. it is known that the dead load for such asix inch slab is 75 lbs. per square foot, so that the total unit floorloading is lbs. per square foot. Thus the total load on any one columnis given by 175 20 20 that is 70,000 lbs. The thickness t is equal to 6inches so that d, the effective depth, may be taken as 61.5=4.5 inches.For a 3000 lbs/sq. in., 28 day strength concrete .03 is is equal to 90lbs. per square inch and as we have seen above, j is taken as beingequal to .875. Therefore, We may write a i.e. 49.4 inches It isinteresting to compare this reduced value of shear stress along theperimeter of a square 2 concentric with the column with the maximumvalue of shear stress which would prevail in the absence ofreinforcement. This maximum value of shear stress vC is given by thefollowing equation:

where b is the specific perimeter of the median of the primary shearstress pattern and is equal to 4 (c-t-d) where c is the side length ofthe square column. In the example cited above, the column had a side of14 inches so that:

Thus it will be seen, that the maximum shear stress in the floor has tobe reduced by approximately 150 lbs. per square inch and to do this itis necessary to provide shear reinforcement extending outwardly to theedges of a square area of side approximately 50 ins. concentricallyabout the column to ensure that this reduction is accomplished.

A graphical representation of the significance of the above calculationsis shown in FIGURE 8 in which is depicted a square slab of the floor ofside p supported upon a square column of side c which has been raisedfrom the floor. The reinforcing steel T located at a distance d abovethe lower surface of the slab floor, the thickness of the slabs being I.

As well as computing the extent over which shear reinforcement isnecessary it is also a requirement to determine how many inclinedreinforcement planes are needed to intersect the successive planes ofmaximum shear stress, and a graphical representation of how this is doneis shown in FIGURE 4, here the reinforcement extends over a distance d,which as shown above is slightly less than t, the thickness of thefloor. The shear stress perimeter is considered as being at the midpoint of the shear line i.e. where this line intersects the inclinedportion of the reinforcement bar, and so the distance between opposedintersection points which was designated earlier as p in the case of thesquare sided column, will be measured along the median of the floor oralong a line running through the floor at a depth d/2 above or below thereinforcing steel.

From FIGURE 4 it will be seen that, for any given floor configuration,the primary plane of maximum shear stress must intersect the reinforcingbar at a fixed distance from the column of d/2, cot 45 i.e. d/2, thetotal span along the median between opposed bars intersecting theprimary plane of maximum shear thus being approximately c-l-d, andaccordingly independent of the angle of inclination of the inclinedportion of the that.

A further distance beyond this primary span is however very muchdependent upon 0 and for one side this in- =240 lbs/sq. in.

I number, and must be rationalixed to a whole number.

creased distance is given by d/2 cot +d/2, the total distance betweenopposed bars should thus be (c+d) +2.(g+g cot 9) (c+d)+1.5(g+g m a) the(c+d) quantity has not been reduced since this factor is itself alreadya reduced approximation from its geometric value of (c+t), and isreduced thereby in approximately the same order of magnitude.

For N reinforcement planes this distance becomes and this must beequaled to the distance between opposed reinforcement planes deduced asabove from the computed value of the perimeter i.e. for a square sidedcolumn It should be noted that reducing the value of 0 increases thedistance between reinforcement planes, and so its value is preferablylow.

A reticulated structure formed as a preferred embodiment according tothe disclosure of the present invention is shown in FIGURES 9, 10 and lland provides the necessary area reinforcement to enable the shear stresspresent in a flat slab floor system to be borne without the use of droppanels or column capitals.- It will be seen that the structure issymmetrical in a horizontal plane about 2 axes at right angles to oneanother and comprises a first of mutually parallel bars across which islaid a second of mutually parallel bars substantially at right angles tothe first. Each of these bars is formed in the manner shown in FIGURE 2aand has a flat central portion 1 on either side of which is disposed adownwardly and outwardly inclined portion at an angle of inclination of1806 to the horizontal, outwardly beyond which the bar has two furtherspaced horizontal portions 3..

The bar generally terminates as shown at opposite ends in an anchoringhook 4 which insures thatit is firmly imbedded in the concrete. Suchhooks are well known and do not form an essential feature of the presentinvention. By varying the length of the centre section 1, it is possibleto move the opposed inclined portions 2 varying distances apart and thusby combining in each of the sets of bars a number of bars whosecentrallength is varied, it is possible to provide reinforcing in aseries of outwardly disposed inclined planes which lie across thesuccessive planes of maximum shear stress and thus move the shear lineoutwardly from the column. The angle 0 may usefully lie anywhere in therange from 15 to 60 degrees but is preferably in the region of twentydegrees since this has the dual benefit of increasing the extent towhich the shear plane is offset by each inclined plane and also re--duces the crushing stress which appears at the junctions of the inclinedportions with two flat portions.

The manner in which the bars of various lengths are interspersed, ineach of two series of bars at right angles to one another and the totalnumber of bars in each. series, is of course, dependent on theparticular circum-.

stances of each case. However, since it is required to defineprogressively wider inclined areas of reinforcement, it follows that theoutermost bars of each set will be progressively spaced further apartwith the outermost.

'bars of the series having the longest centre section and that the barsintermediate to these outer bars will be: varied as to number and lengthof centre sections in con-. formity with the need for providingsufficient reinforcement in each inclined plane and for distributing thereinforcement substantially uniformly in each inclined plane.

The shear head shown in FIGURES 9 and 10 consists of a first and secondset of members each set being in:

a grid like pattern. The first set consists of a first series of bars11a, b, c, d, and e and a second series 13a, b, c,

d and e disposed at right angles thereto. Bars 11c and respectively formthe respective axis ofa quadrature and bars 11a and b; 11b and e; 13aand b and 130 and d are respectively disposed symmetrically on eitherside thereof. The inclined portions of bars 11 and 13 define a fiirstinclined plane of reinforcement disposed transverse.

to the axis thereof.

The second set consists of similarly arranged and 12 having likealphabetical reference indicating a corresponding arrangement andforming respectively a first and second series. symmetrically placed oneither side of the axis of the first series in the first set arev longerbars 10a, 10b, 10c and 10d forming a first series in the second set andwhose inclined portions define a second inclined plane of reinforcementsymmetrically spaced outwardly from the first inclined plane provided bythe shorter set of bars 11 above. Bars 10a and 10d bars 10 form the twoouter bars of this first series in the second set and bars 10b and 100are disposed symmetrically inwardly of these two outer bars and areinterposed between the shorter bars with bar 10b being positionedbetween shorter bars 11b and 11c of the first set and bar 100 beinglocated between bars 11c and 11d.

A second series of bars in the second set consisting of bars 12a b, cand d are disposed symmetrically at right angles to the first series ofthe second set and are similarly arranged with respect to bars 13a, 13c,13d, 132.

The diameter of the bars is determined from a consideration of a numberof bars in the particular inclined reinforcement surface and the amountof reinforcement needed in this surface to resist, with an adequatemargin of safety, the shear force on the surface, which may be suitablyand conveniently expressed in square inches, in cross-section, of steelper surface. Thus in a specific shearhead designed according to thisinvention, it was found necessary to have 0.95 square inch of steelreinforcement in the first inclined surface, and 0.65 square inch in thesecond inclined surface, and, using the configuration described aboveemploying five bars in the first set and four bars in the second set,these two amounts were provided by five A2 inch diameter bars supplyingapproximately a total of one square inch, and four /2 inch diameter barssupplying approximately 0.80 square inch respectively for the twosurfaces.

This particular combination of number and size bars was chosen for theirconvenient approximation of the required reinforcement and because theirresultant spacing seemed preferable, but equally, the first surfacewould have been reinforced with three inch diameter bars and the secondsurface with six /8 inch diameter bars. The number and size of the barsis therefore moderately flexible, the primary requirement being that, inthe aggregate, they provide adequate shear reinforcement.

The first series of bars in each set are secured together at oppositeends by a tie :bar; the two tie bars in the case of the first serieswhich comprises shorter bars 11 and long bars 10 being tie bars 14 and15 and the corresponding tie bars for the second series being tie bars16 and 17. These tie bars each have at either end a small verticalportion which depends downwardly below the general lower level of theshear-head and ensures that the inclined reinforcing surfaces of theshearhead are correctly positioned vertically with respect to the shearlines.

The shear reinforcement structure described above can be formed bypositioning the bars in their correct respective stations actually onthe job. However, this task is both time consuming and requires anundesirably high level of skill on the part of the workmen forming thestructure and it is one of the virtues of the sheanhead structure formedaccording to the present invention that it can be prefabricatedelsewhere in a suitable workshop employing jigs and braces of theappropriate size and shape, the bars either being tied or spot weldedtogether when correctly positioned. The shearheads can be prefabricatedand stored until required and then positioned on the job with a minimumof time and effort by relatively unskilled workmen.

FIGURE 11 shows a skeleton view of this shearhead structure M as itwould appear when placed above a column C in a fiat plate fioor F. Whilethe parts are the same as in FIGURES 9 and 10, the numerals are omittedfor greater clarity and simplicity.

It is necessary to modify this structure when the column is locatedatone of the outer walls of the building and a typical manner in whichthis structure is adapted is shown in FIGURE 12, though this must bedone without disrupting the requirement that the shear reinforcement beprovided over a unitary inclined surface of reinforcement symmetric withrespect to the column in all directions over which shear reinforcementis necessary.

Since the column is at the outer wall, no shear load is presented to itfrom beyond the wall, the full shear load is presented from a directionnormal to the wall and a lesser shear load approximately half the fullload is presented from direction on either side of the column parallelto the wall. Thus it logically follows that the number of bars in thatseries parallel to the wall may :be halved without any appreciable lossin shear reinforcement. Accordingly, those bars in the series parallelto the wall which would normally lie outwardly beyond this wall areremoved and the ends of the bars [forming the series normal to the wallare bent downward through degrees until they lie in the thickened endportion W of the flat plate floor F above column C.

I claim:

A reinforced concrete structure comprising a flat plate floor having anupper and lower surface, at least one supporting column, and shearreinforcement embodied in said concrete adjacent said column, saidreinforcement comprising a shear head having a first set of reinforcingmembers consisting of a first series of bars arranged substantiallyparallel in equally spaced side-by side relationship and a second seriesof similarly arranged bars at right angles thereto, the bars in eachseries having an upper horizontal central portion substantially parallelto and adjacent the upper surface of said floor and opposed horizontalanchoring portions interconnected to said central portion by a pair ofdownwardly, outwardly inclined portions said inclined portions of eachseries lying in a plane angula-rly disposed to the axis of therespective series to present a first plane of reinforcement and a secondset of similarly arranged reinforcing members consisting of a first andsecond series of bars interspersed symmetrically with the respectiveseries of said first set and having similarly inclined portions topresent a second plane of reinforcement, said planes of reinforcementbeing located symmetrical with respect to said column, said second setof bars having an upper horizontal central portion of greater lengththan that of the first set whereby said second plane of reinforcement islocated more remote from said column than said first set thereby toprovide reinforcement respectively to intersect secondary and primaryshear stresses in said floor adjacent said supporting column.

References Cited by the Examiner UNITED STATES PATENTS 729,299 5/ 1903Ellinger et a1. 52252 896,963 8/ 1908 White 52721 938,393 10/1909 Martin52251 1,033,797 7/1912 Hartman 52252 1,052,708 2/ 1913 Anderson 522601,072,532 9/1913 'Iurner 52251 1,078,510 11/1913 Luten 52252 1,119,40612/1914 Danielson 52260 1,143,527 6/1915 Francis 52250 1,163,853 12/1915Randall et a1. 52264 1,182,421 5/1916 Ramsey 52251 1,205,347 11/ 1916Hincz 52252 1,244,641 10/1917 Pratt 5225 1 FOREIGN PATENTS 17,179 12/1929 Australia. 326,638 11/ 1902 France. 498,886 9/ 1954 Italy.

FRANK L. ABBOTT, Primary Examiner.

HENRY C. SUTHERLAND, Examiner.

I. L. RIDGILL, Assistant Examiner.

